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Tensile forces stretch an object.
Compressive forces squash an object.

","mcqOptions":[{"text":"

Tensile forces stretch an object.
Compressive forces squash an object.

","correct":true,"explanation":"Tensile forces pull parts of an object apart, increasing its length, whereas compressive forces push parts together and reduce its length."},{"text":"

Tensile forces squash an object.
Compressive forces stretch an object.

","correct":false,"explanation":"This reverses the two force types. Tensile forces pull and stretch an object, while compressive forces push inward and squash it."},{"text":"

Tensile forces bend an object.
Compressive forces twist an object.

","correct":false,"explanation":"Bending and twisting describe different forms of deformation, not tensile and compressive loading. Tensile forces stretch an object, while compressive forces squash it."},{"text":"

Tensile forces stretch an object.
Compressive forces stretch an object.

","correct":false,"explanation":"Both forces are incorrectly described as stretching. Tensile forces stretch an object, but compressive forces act in the opposite direction and squash it."}]},{"id":"9f9860e5-dd32-4b55-85c4-c2147b6efe93","cardType":"cloze","points":null,"front":"To calculate weight in newtons, the mass must be in $\\text{[...]}$ and multiplied by $g = 9.81\\ \\text{N kg}^{-1}$.","back":"To calculate weight in newtons, the mass must be in $\\boxed{\\text{kg}}$ and multiplied by $g = 9.81\\ \\text{N kg}^{-1}$.","mcqOptions":[{"text":"$$\\text{kg}$","correct":true,"explanation":"The equation $W = mg$ requires mass in kilograms because gravitational field strength is measured in newtons per kilogram, giving weight in newtons."},{"text":"$$\\text{g}$","correct":false,"explanation":"Using grams without conversion makes the numerical result 1000 times too large. Mass must first be expressed in kilograms when using $g = 9.81\\ \\text{N kg}^{-1}$."},{"text":"$$\\text{N}$","correct":false,"explanation":"Newtons measure force rather than mass, so they cannot be used for the mass in $W = mg$. The required mass unit is kilograms."},{"text":"$$\\text{lb}$","correct":false,"explanation":"Pounds are not the SI unit of mass and are incompatible with the stated value of $g$. The mass must be converted to kilograms."}]},{"id":"741f9a62-2046-410a-a943-d5b3b3d971cb","cardType":"cloze","points":null,"front":"To calculate weight in newtons, the mass must be in $\\text{kg}$ and multiplied by $\\text{[...]}$.","back":"To calculate weight in newtons, the mass must be in $\\text{kg}$ and multiplied by $\\boxed{g = 9.81\\ \\text{N kg}^{-1}}$.","mcqOptions":[{"text":"$$g = 9.81\\ \\text{N kg}^{-1}$","correct":true,"explanation":"Gravitational field strength near Earth’s surface is approximately $9.81\\ \\text{N kg}^{-1}$. Multiplying this by mass in kilograms gives weight in newtons."},{"text":"$$g = 0.981\\ \\text{N kg}^{-1}$","correct":false,"explanation":"This decimal-place slip makes gravitational field strength ten times too small, so the calculated weight is one tenth of its correct value. Use $g = 9.81\\ \\text{N kg}^{-1}$."},{"text":"$$g = 9.81\\ \\text{N m}^{-1}$","correct":false,"explanation":"This uses the spring constant’s units rather than gravitational field strength’s units. Weight is found using $g = 9.81\\ \\text{N kg}^{-1}$."},{"text":"$$g = 9.81\\ \\text{kg N}^{-1}$","correct":false,"explanation":"This reverses the units of gravitational field strength. To obtain newtons after multiplying by kilograms, use $g = 9.81\\ \\text{N kg}^{-1}$."}]},{"id":"dd541045-cd5c-47d8-bd5c-d9f0ebd13199","cardType":"basic","points":1,"front":"How is extension ($x$) calculated from original length ($L_0$) and stretched length ($L$)?","back":"$$$x = L - L_0$$","mcqOptions":[{"text":"$$$x = L - L_0$$","correct":true,"explanation":"Extension is the increase in length, so it equals the final stretched length minus the original length. Both lengths must use the same units."},{"text":"$$$x = L_0 - L$$","correct":false,"explanation":"Reversing the subtraction gives a negative value when the object is stretched. Extension is the stretched length minus the original length: $x = L - L_0$."},{"text":"$$$x = L + L_0$$","correct":false,"explanation":"Adding the two lengths does not measure the increase in length. Extension is the difference between final and original lengths: $x = L - L_0$."},{"text":"$$$x = \\frac{L}{L_0}$$","correct":false,"explanation":"Dividing the lengths gives a dimensionless ratio rather than an extension with units of length. The extension is calculated using $x = L - L_0$."}]},{"id":"0738cb54-1c4a-4b65-b260-683f461de714","cardType":"basic","points":2,"front":"State Hooke’s law.","back":"The force applied is directly proportional to the extension, provided the limit of proportionality is not exceeded.","mcqOptions":[{"text":"The force applied is directly proportional to the extension, provided the limit of proportionality is not exceeded.","correct":true,"explanation":"Within the limit of proportionality, doubling the applied force doubles the extension, so force and extension have a constant ratio represented by the stiffness."},{"text":"The force applied is inversely proportional to the extension, provided the limit of proportionality is not exceeded.","correct":false,"explanation":"This reverses the relationship in Hooke’s law. Before the limit of proportionality, force is directly proportional to extension, so increasing force increases extension proportionally."},{"text":"The force applied is directly proportional to the original length, provided the limit of proportionality is not exceeded.","correct":false,"explanation":"Hooke’s law relates applied force to the change in length, not the object’s original length. Force is directly proportional to extension below the limit of proportionality."},{"text":"The force applied is directly proportional to the extension, provided the limit of proportionality has been exceeded.","correct":false,"explanation":"The proportional relationship does not continue after the limit of proportionality. Hooke’s law applies only while that limit has not been exceeded."}]},{"id":"1463d85f-f51b-4cac-b438-ab7be9f30157","cardType":"cloze","points":null,"front":"In the equation $F = kx$, $k$ represents the [ ] and its units are $\\text{N m}^{-1}$.","back":"In the equation $F = kx$, $k$ represents the stiffness (or spring constant) and its units are $\\text{N m}^{-1}$.","mcqOptions":[{"text":"stiffness (or spring constant)","correct":true,"explanation":"The constant $k$ is the spring’s stiffness, also called its spring constant. It is the force required per unit extension in $F = kx$."},{"text":"extension (or change in length)","correct":false,"explanation":"Extension is represented by $x$, not $k$. In $F = kx$, $k$ is the stiffness or spring constant, which measures force required per unit extension."},{"text":"force (or applied load)","correct":false,"explanation":"Force is represented by $F$, not $k$. The symbol $k$ denotes the stiffness or spring constant and has units of newtons per metre."},{"text":"energy (or elastic potential energy)","correct":false,"explanation":"Elastic potential energy is calculated from force and extension but is not represented by $k$. Here, $k$ is the stiffness or spring constant."}]},{"id":"7ad3405d-f33f-4705-9710-d7cfa5c50703","cardType":"cloze","points":null,"front":"In the equation $F = kx$, $k$ represents the stiffness (or spring constant) and its units are $\\text{[...]}$.","back":"In the equation $F = kx$, $k$ represents the stiffness (or spring constant) and its units are $\\boxed{\\text{N m}^{-1}}$.","mcqOptions":[{"text":"$$\\text{N m}^{-1}$","correct":true,"explanation":"Since $k = F/x$, its units are force divided by extension: newtons divided by metres, written as $\\text{N m}^{-1}$."},{"text":"$$\\text{N}$","correct":false,"explanation":"Newtons measure force alone and omit the division by extension. Because $k = F/x$, stiffness has units of newtons per metre, $\\text{N m}^{-1}$."},{"text":"$$\\text{N m}$","correct":false,"explanation":"Newton metres are units of energy or moment, arising from multiplying force by distance. Stiffness divides force by extension, giving $\\text{N m}^{-1}$."},{"text":"$$\\text{m N}^{-1}$","correct":false,"explanation":"Metres per newton describe compliance, the reciprocal of stiffness. Stiffness is force per extension, so its units are $\\text{N m}^{-1}$."}]},{"id":"457156c9-dd65-4157-bf96-113998b96e86","cardType":"basic","points":1,"front":"What does the stiffness ($k$) of a spring indicate?","back":"The amount of force required to produce 1 metre of extension.","mcqOptions":[{"text":"The amount of force required to produce 1 metre of extension.","correct":true,"explanation":"Stiffness is defined by $k = F/x$, so its numerical value gives the force needed for one metre of extension within the proportional region."},{"text":"The amount of extension produced by applying 1 newton of force.","correct":false,"explanation":"This describes extension per unit force, which is compliance and equals $1/k$. Stiffness instead gives the force required to produce one metre of extension."},{"text":"The amount of force required to produce 1 metre of original length.","correct":false,"explanation":"Stiffness concerns force per unit extension, not the object’s original length. It indicates the force required to produce one metre of extension."},{"text":"The amount of energy required to produce 1 metre of extension.","correct":false,"explanation":"Energy required per metre is not the definition of stiffness, since elastic energy varies with the square of extension. Stiffness is force required per metre of extension."}]},{"id":"6db10c3d-f36a-4858-bf6b-4c4793d3edf3","cardType":"basic","points":1,"front":"How do you find the stiffness ($k$) from a graph of Force (y-axis) against extension (x-axis)?","back":"Calculate the gradient of the straight-line section.","mcqOptions":[{"text":"Calculate the gradient of the straight-line section.","correct":true,"explanation":"For a Force-extension graph, Hooke’s law gives $F = kx$, so the gradient $rac{\\Delta F}{\\Delta x}$ is the stiffness $k$."},{"text":"Calculate the gradient of the curved section.","correct":false,"explanation":"The curved section does not have a constant gradient, so it cannot directly give the stiffness. Use the gradient of the straight-line section, where Hooke’s law applies."},{"text":"Calculate the reciprocal of the gradient of the straight-line section.","correct":false,"explanation":"Taking the reciprocal gives the gradient of an extension-Force graph, not the stiffness on a Force-extension graph. Here, stiffness is directly equal to the gradient."},{"text":"Calculate the area under the straight-line section.","correct":false,"explanation":"The area under a Force-extension graph gives elastic strain energy, not stiffness. Stiffness is found from the gradient of the straight-line section."}]},{"id":"b1d9b648-c6af-460c-904a-880c072d9a45","cardType":"basic","points":1,"front":"What does the gradient represent on an Extension-Force graph (extension on y-axis, force on x-axis)?","back":"$$$\\frac{1}{k}$$ (the reciprocal of the stiffness)","mcqOptions":[{"text":"$$$\\frac{1}{k}$$ (the reciprocal of the stiffness)","correct":true,"explanation":"With extension on the y-axis and force on the x-axis, the gradient is $rac{\\Delta x}{\\Delta F}$. Since $F = kx$, this equals $rac{1}{k}$."},{"text":"$$$k$$ (the stiffness)","correct":false,"explanation":"This is the gradient when Force is plotted against extension. Reversing the axes gives $rac{\\Delta x}{\\Delta F} = \\frac{1}{k}$, the reciprocal of stiffness."},{"text":"$$$k^2$$ (the square of the stiffness)","correct":false,"explanation":"Squaring the stiffness has no basis in the graph’s gradient. Reversing the Force and extension axes takes the reciprocal, giving $rac{1}{k}$."},{"text":"$$$\\frac{1}{k^2}$$ (the reciprocal of the square of the stiffness)","correct":false,"explanation":"The gradient involves one change in extension divided by one change in force, not squared quantities. Therefore it is $rac{1}{k}$, not $rac{1}{k^2}$."}]},{"id":"3709037b-a60c-4033-bb07-32a183626ca7","cardType":"basic","points":1,"front":"Define the limit of proportionality.","back":"The point beyond which force is no longer directly proportional to extension.","mcqOptions":[{"text":"The point beyond which force is no longer directly proportional to extension.","correct":true,"explanation":"The limit of proportionality is reached when the Force-extension graph stops being linear. Beyond this point, force and extension are no longer directly proportional."},{"text":"The point beyond which force no longer causes any extension in the object.","correct":false,"explanation":"The object generally continues extending beyond this point, but extension is no longer proportional to force. The correct definition concerns the end of the linear relationship."},{"text":"The point beyond which extension is no longer directly proportional to stiffness.","correct":false,"explanation":"Stiffness is the gradient of the linear region, not the variable that extension is directly proportional to. The limit concerns the relationship between force and extension."},{"text":"The point beyond which force causes permanent deformation of the object.","correct":false,"explanation":"Permanent deformation begins at the elastic limit, which can occur after the limit of proportionality. The limit of proportionality only marks the end of linear behaviour."}]},{"id":"f69b31da-423b-4e98-8ccd-1a4f7ac19105","cardType":"basic","points":1,"front":"Define elastic deformation.","back":"The object returns to its original shape and length when the force is removed.","mcqOptions":[{"text":"The object returns to its original shape and length when the force is removed.","correct":true,"explanation":"Elastic deformation is reversible: removing the force allows the object to recover its original shape and length, provided its elastic limit has not been exceeded."},{"text":"The object keeps its changed shape and length when the force is removed.","correct":false,"explanation":"Keeping a changed shape describes plastic deformation, not elastic deformation. Elastic deformation is reversible, so the object returns to its original shape and length."},{"text":"The object returns to its original shape but not length when the force is removed.","correct":false,"explanation":"A fully elastic object returns to both its original shape and length. A remaining change in length indicates permanent, plastic deformation."},{"text":"The object breaks into separate pieces when the force is removed.","correct":false,"explanation":"Breaking is fracture, not elastic deformation. Elastic deformation involves a temporary change that reverses when the applied force is removed."}]},{"id":"a7a38004-330d-45a0-b5ee-dd5f3f6ac445","cardType":"basic","points":1,"front":"Define plastic deformation.","back":"The object is permanently changed and does not return to its original length after unloading.","mcqOptions":[{"text":"The object is permanently changed and does not return to its original length after unloading.","correct":true,"explanation":"Plastic deformation is irreversible. Once the force is removed, the object retains a permanent change in length or shape instead of returning fully to its original form."},{"text":"The object is temporarily changed and returns to its original length after unloading.","correct":false,"explanation":"A temporary change that reverses after unloading is elastic deformation. Plastic deformation leaves a permanent change in the object."},{"text":"The object is permanently changed but returns to its original length after unloading.","correct":false,"explanation":"Returning to the original length contradicts permanence and describes elastic behaviour. Plastic deformation leaves the object with a permanent extension or altered shape."},{"text":"The object is unchanged and does not extend under an applied force.","correct":false,"explanation":"Plastic deformation requires a permanent change caused by force. An unchanged object has not undergone deformation, whether elastic or plastic."}]},{"id":"934ad337-49e1-4046-8b59-b93d9f2e5e28","cardType":"basic","points":1,"front":"What is the elastic limit?","back":"The maximum force an object can experience and still return to its original length.","mcqOptions":[{"text":"The maximum force an object can experience and still return to its original length.","correct":true,"explanation":"The elastic limit is the greatest force for which deformation remains fully reversible. Applying a greater force causes permanent deformation when the object is unloaded."},{"text":"The minimum force an object needs to experience before it begins to extend.","correct":false,"explanation":"This describes a possible threshold for starting extension, not the elastic limit. The elastic limit is the maximum force that still permits complete recovery."},{"text":"The maximum force an object can experience before it breaks apart.","correct":false,"explanation":"The force causing fracture is the breaking force, not the elastic limit. Permanent deformation may begin well before the object actually breaks."},{"text":"The maximum extension an object can reach and still remain directly proportional to force.","correct":false,"explanation":"This confuses extension and force and describes proportional behaviour. The elastic limit is defined by whether the object returns to its original length after unloading."}]},{"id":"c0f049ef-262e-4ac4-919e-d0f7ffb4d360","cardType":"basic","points":2,"front":"How do the limit of proportionality and elastic limit differ?","back":"

Limit of proportionality: where the linear relationship ends.
Elastic limit: where permanent deformation begins.

","mcqOptions":[{"text":"

Limit of proportionality: where the linear relationship ends.
Elastic limit: where permanent deformation begins.

","correct":true,"explanation":"The limit of proportionality marks the end of the straight-line Force-extension relationship. The elastic limit marks the onset of permanent deformation, and it may occur later."},{"text":"

Limit of proportionality: where permanent deformation begins.
Elastic limit: where the linear relationship ends.

","correct":false,"explanation":"These definitions have been swapped. The limit of proportionality ends linear behaviour, whereas the elastic limit is where deformation becomes permanent."},{"text":"

Limit of proportionality: where the object breaks.
Elastic limit: where extension begins.

","correct":false,"explanation":"Neither definition is correct: extension can begin at much smaller forces, and breaking occurs after substantial deformation. The two limits concern linearity and reversibility."},{"text":"

Limit of proportionality: where stiffness becomes zero.
Elastic limit: where the object reaches maximum extension.

","correct":false,"explanation":"The limit of proportionality does not require zero stiffness, and the elastic limit is not defined by maximum extension. They concern the linear relationship and permanent deformation."}]},{"id":"c02ad642-2e39-469b-b4fe-97000ec56a4b","cardType":"basic","points":1,"front":"How is elastic strain energy found from a force-extension graph?","back":"Calculate the area under the graph.","mcqOptions":[{"text":"Calculate the area under the graph.","correct":true,"explanation":"Elastic strain energy is the work done stretching the object. On a force-extension graph, work done equals the area under the graph."},{"text":"Calculate the gradient of the graph.","correct":false,"explanation":"The gradient of a Force-extension graph gives stiffness in the linear region. Elastic strain energy is found from the area under the graph."},{"text":"Calculate the area above the graph.","correct":false,"explanation":"The work done is represented by the area between the graph and the extension axis, not the area above the graph. That area gives elastic strain energy."},{"text":"Calculate the reciprocal of the area under the graph.","correct":false,"explanation":"Taking a reciprocal is not part of calculating work done. The area under the force-extension graph directly represents the elastic strain energy stored."}]},{"id":"00567abc-2306-4d7e-8cde-8d351ee4f20a","cardType":"cloze","points":null,"front":"For a spring obeying Hooke's law, the work done stretching it is given by $E = \\text{[...]}$ or $E = \\frac{1}{2}kx^2$.","back":"For a spring obeying Hooke's law, the work done stretching it is given by $E = \\boxed{\\frac{1}{2}Fx}$ or $E = \\frac{1}{2}kx^2$.","mcqOptions":[{"text":"$$\\frac{1}{2}Fx$","correct":true,"explanation":"The work done is the area beneath the linear force-extension graph. This triangular area is half the final force multiplied by the extension, giving $E = \\frac{1}{2}Fx$."},{"text":"$$Fx$","correct":false,"explanation":"Multiplying $F$ by $x$ treats the force as constant at its final value. The force rises from zero to $F$, so its average is $\\frac{1}{2}F$ and $E = \\frac{1}{2}Fx$."},{"text":"$$\\frac{1}{2}Fx^2$","correct":false,"explanation":"Squaring the extension confuses this form with $\\frac{1}{2}kx^2$. When final force and extension are used, the triangular area gives $E = \\frac{1}{2}Fx$."},{"text":"$$\\frac{1}{2}kx$","correct":false,"explanation":"This omits one factor of extension and therefore does not have units of energy. Using $F = kx$ in the triangular-area expression gives $E = \\frac{1}{2}Fx$."}]},{"id":"7abc4aac-4e90-4cf9-ac78-9fdab17388ad","cardType":"cloze","points":null,"front":"For a spring obeying Hooke's law, the work done stretching it is given by $E = \\frac{1}{2}Fx$ or $E = \\text{[...]}$.","back":"For a spring obeying Hooke's law, the work done stretching it is given by $E = \\frac{1}{2}Fx$ or $E = \\boxed{\\frac{1}{2}kx^2}$.","mcqOptions":[{"text":"$$\\frac{1}{2}kx^2$","correct":true,"explanation":"Since Hooke's law gives $F = kx$, substituting into $E = \\frac{1}{2}Fx$ produces $E = \\frac{1}{2}kx^2$, the elastic strain energy stored."},{"text":"$$kx^2$","correct":false,"explanation":"This misses the factor of one-half by treating the final force as acting throughout the extension. The force increases linearly from zero, so $E = \\frac{1}{2}kx^2$."},{"text":"$$\\frac{1}{2}kx$","correct":false,"explanation":"This uses only one factor of extension, giving units of force rather than energy. Substituting $F = kx$ into $E = \\frac{1}{2}Fx$ gives $E = \\frac{1}{2}kx^2$."},{"text":"$$\\frac{1}{2}k^2x$","correct":false,"explanation":"Squaring stiffness instead of extension results from substituting Hooke's law incorrectly. Because $F = kx$, the correct substitution gives $E = \\frac{1}{2}kx^2$."}]},{"id":"deacecca-0e12-4c97-b690-629cfc4841bc","cardType":"basic","points":1,"front":"When is the equation $E = \\frac{1}{2}Fx$ not valid for calculating energy stored?","back":"When the force-extension graph is not a straight line through the origin.","mcqOptions":[{"text":"When the force-extension graph is not a straight line through the origin.","correct":true,"explanation":"The formula uses the triangular area beneath a linear force-extension graph. If the graph is not straight through the origin, its area is not generally $\\frac{1}{2}Fx$."},{"text":"When the force-extension graph is a straight line through the origin.","correct":false,"explanation":"A straight line through the origin is precisely the condition under which the area beneath the graph is triangular. The equation fails when the graph is not linear through the origin."},{"text":"When the force-extension graph has a constant gradient through the origin.","correct":false,"explanation":"A constant gradient through the origin means force is proportional to extension, so $\\frac{1}{2}Fx$ is valid. It becomes invalid when that linear relationship does not hold."},{"text":"When the force-extension graph shows Hooke's law up to the final extension.","correct":false,"explanation":"Obeying Hooke's law makes the graph linear through the origin, so its area is $\\frac{1}{2}Fx$. The equation is not valid when the graph is non-linear."}]},{"id":"2ccbd482-e48e-43c5-94e6-9729fc2ef0a4","cardType":"cloze","points":null,"front":"Elastic strain energy is measured in the SI unit [ ].","back":"Elastic strain energy is measured in the SI unit joules (J).","mcqOptions":[{"text":"joules (J)","correct":true,"explanation":"Elastic strain energy is a form of energy, and the SI unit for every form of energy and work is the joule, with unit symbol J."},{"text":"newtons (N)","correct":false,"explanation":"Newtons measure force, which may be used to calculate elastic strain energy but is not its unit. Energy is measured in joules, with unit symbol J."},{"text":"watts (W)","correct":false,"explanation":"Watts measure power, meaning the rate of energy transfer, rather than energy itself. Elastic strain energy is therefore measured in joules, with unit symbol J."},{"text":"pascals (Pa)","correct":false,"explanation":"Pascals measure pressure or stress, not stored energy. Although stress is relevant to deformation, elastic strain energy is measured in joules, with unit symbol J."}]},{"id":"4cf98636-f553-4f00-9901-ab7b44011484","cardType":"basic","points":2,"front":"How can parallax error be reduced when measuring the length of a spring?","back":"

Read the scale at eye level.
Use a fiducial marker or pointer.

","mcqOptions":[{"text":"

Read the scale at eye level.
Use a fiducial marker or pointer.

","correct":true,"explanation":"Viewing at eye level makes the line of sight perpendicular to the scale, while a fiducial marker defines the reading position clearly. Both measures reduce parallax error."},{"text":"

Read the scale from directly above.
Use a wide marker or pointer.

","correct":false,"explanation":"Viewing from above creates an angled line of sight, while a wide marker makes the reading position ambiguous. Read at eye level and use a narrow fiducial marker or pointer."},{"text":"

Read the scale at an angle.
Use the spring's nearest edge as a pointer.

","correct":false,"explanation":"An angled view causes parallax, and the spring's edge may not define a precise reference point. Read at eye level using a clear fiducial marker or pointer."},{"text":"

Place the ruler behind the spring.
View the scale from either side.

","correct":false,"explanation":"Placing the scale and spring in different planes and viewing from the side increases parallax. The scale should be read at eye level using a fiducial marker or pointer."}]},{"id":"94f68e2b-f2f9-4896-b892-9215a293f406","cardType":"basic","points":2,"front":"If a ruler has an uncertainty of $\\pm 1\\ \\text{mm}$, why is the uncertainty in extension $\\pm 2\\ \\text{mm}$?","back":"Extension is the difference between two readings (stretched length and original length), so the uncertainties are added.","mcqOptions":[{"text":"Extension is the difference between two readings (stretched length and original length), so the uncertainties are added.","correct":true,"explanation":"Extension is calculated by subtracting two measured lengths. For worst-case absolute uncertainty in a subtraction, the two absolute uncertainties add, giving $1\\ \\text{mm} + 1\\ \\text{mm} = 2\\ \\text{mm}$."},{"text":"Extension is the difference between two readings (stretched length and original length), so the uncertainties are subtracted.","correct":false,"explanation":"Subtracting measured values does not mean their uncertainties are subtracted. Worst-case absolute uncertainties add for both addition and subtraction, so the extension uncertainty is $\\pm 2\\ \\text{mm}$."},{"text":"Extension is the sum of two readings (stretched length and original length), so the uncertainties are multiplied.","correct":false,"explanation":"Extension is found by subtracting the original length from the stretched length, not adding them. The two absolute uncertainties are then added, producing an uncertainty of $\\pm 2\\ \\text{mm}$."},{"text":"Extension is a single ruler reading (the stretched length only), so the uncertainty is unchanged.","correct":false,"explanation":"Extension requires both the stretched and original lengths, rather than only one reading. Each contributes $\\pm 1\\ \\text{mm}$, so their absolute uncertainties add to $\\pm 2\\ \\text{mm}$."}]},{"id":"05be3c7a-0fa3-467c-a0f3-a35e8692f5c4","cardType":"basic","points":1,"front":"Why is stiffness ($k$) not a property of the material alone?","back":"It also depends on the object's shape and dimensions (e.g., a wire vs a spring of the same material).","mcqOptions":[{"text":"It also depends on the object's shape and dimensions (e.g., a wire vs a spring of the same material).","correct":true,"explanation":"Stiffness describes the response of a particular object, so changing its length, cross-sectional area or geometry changes $k$ even when the material remains the same."},{"text":"It depends only on the material's density and mass (e.g., heavy vs light objects made from the same material).","correct":false,"explanation":"Mass and density do not alone determine an object's stiffness. Stiffness depends on material properties together with the object's shape and dimensions, such as length, cross-sectional area and geometry."},{"text":"It also depends on the applied force and extension (e.g., using a larger force gives a larger stiffness).","correct":false,"explanation":"Within the Hooke's law region, increasing force produces a proportional extension, so the ratio $F/x$ remains constant. Stiffness instead depends on material, shape and dimensions."},{"text":"It depends only on the object's weight and orientation (e.g., vertical vs horizontal use of the same object).","correct":false,"explanation":"Weight and orientation do not define intrinsic stiffness in the Hooke's law model. The value of $k$ depends on the material together with the object's shape and dimensions."}]}]}],"examCountdown":{"examDate":"2027-05-12","paperLabel":"Paper 1"},"heading":"Hooke's law and stiffness","title":"Hooke's law and stiffness Flashcards","topbar":{"backHref":"/as-level/physics/edexcel/flashcards"},"topicSidebar":"$L37"}]}],"topicId":"df037715-b83b-5f1c-a41d-b4569319ea3d","walled":"$L38"}],"$L39","$L3a"]